derivatives equivalent somehow

Question

If I take the derivative of

$$\frac{1}{1-x}$$

I get:

$$\frac{1}{(1-x)^2}$$

If I take the derivative of

the same as $$\frac{x}{1-x}$$

I also get

$$\frac{1}{(1-x)^2}$$

Am I doing something wrong?

Answer 1

Hint:

$$\frac x{1-x}=-1+\frac1{1-x}$$

Answer 2

Am I doing something wrong?

No, why? Indeed $$ \frac{x}{1-x}=\frac{1-1+x}{1-x}=\frac{1}{1-x}-1 $$ shows that the two functions differ by a constant, so they have the same derivative.

Answer 3

Well, $$\frac{1}{1-x}-\frac{x}{1-x}=1$$ so $$\frac{d}{dx}\frac{1}{1-x}-\frac{d}{dx}\frac{x}{1-x}=\frac{d}{dx}1=0$$